如何从信号中去除频率

我想从信号中删除一个频率（一个峰值）并在没有它的情况下绘制我的函数。在 fft 之后，我找到了频率和幅度，我不确定现在需要做什么。例如，我想删除我的最高峰（​​在绘图上用红点标记）。

import numpy as np
import matplotlib.pyplot as plt

# create data
N = 4097
T = 100.0
t = np.linspace(-T/2,T/2,N)
f = np.sin(50.0 * 2.0*np.pi*t) + 0.5*np.sin(80.0 * 2.0*np.pi*t)

#plot function
plt.plot(t,f,'r')
plt.show()

# perform FT and multiply by dt
dt = t[1]-t[0]
ft = np.fft.fft(f) * dt      
freq = np.fft.fftfreq(N, dt)
freq = freq[:N/2+1]
amplitude = np.abs(ft[:N/2+1])
# plot results
plt.plot(freq, amplitude,'o-')
plt.legend(('numpy fft * dt'), loc='upper right')
plt.xlabel('f')
plt.ylabel('amplitude')
#plt.xlim([0, 1.4])


plt.plot(freq[np.argmax(amplitude)], max(amplitude), 'ro')
print "Amplitude: " + str(max(amplitude)) + "  Frequency: " + str(freq[np.argmax(amplitude)])

plt.show()

一种选择是将信号变换到频域，然后删除选定的频率。

import numpy as np
import matplotlib.pyplot as plt
from scipy.fftpack import rfft, irfft, fftfreq, fft

# Number of samplepoints
N = 500
# sample spacing
T = 0.1

x = np.linspace(0.0, (N-1)*T, N)
# x = np.arange(0.0, N*T, T)  # alternate way to define x
y = 5*np.sin(x) + np.cos(2*np.pi*x) 

yf = fft(y)
xf = np.linspace(0.0, 1.0/(2.0*T), N//2)
#fft end

f_signal = rfft(y)
W = fftfreq(y.size, d=x[1]-x[0])

cut_f_signal = f_signal.copy()
cut_f_signal[(W>0.6)] = 0  # filter all frequencies above 0.6

cut_signal = irfft(cut_f_signal)

# plot results
f, axarr = plt.subplots(1, 3, figsize=(9, 3))
axarr[0].plot(x, y)
axarr[0].plot(x,5*np.sin(x),'g')

axarr[1].plot(xf, 2.0/N * np.abs(yf[:N//2]))
axarr[1].legend(('numpy fft * dt'), loc='upper right')
axarr[1].set_xlabel("f")
axarr[1].set_ylabel("amplitude")


axarr[2].plot(x,cut_signal)
axarr[2].plot(x,5*np.sin(x),'g')

plt.show()